Example

The space $\mathbf{R}^A$ of all real-valued functions defined on any set $A$ is a Riesz space (for the order relation $\langle x(t) \leq y(t) \text{ for all } t \in A \rangle$); for, any two real-valued functions $x, y$ defined on $A$ have a supremum (resp. an infimum) equal to the mapping $t \mapsto \sup(x(t), y(t))$ (resp. $t \mapsto \inf(x(t), y(t))$).

One can also say that a Riesz space is a vector space $E$ equipped with an order structure such that, on the one hand, this structure and the additive group structure of $E$ define a lattice-ordered group structure on $E$ (A, VI, §1, No. 9), and on the other hand that the axiom (OVS$_{\mathrm{II}}$) is satisfied.

Thus, all of the properties of lattice-ordered groups are applicable to Riesz spaces; we shall review here the principal ones (cf. A, VI, §1, Nos. 9 to 12), as well as indicating the consequences that flow from the axiom (OVS$_{\mathrm{II}}$).

We recall first that one writes $x^+ = \sup(x, 0)$ (the positive part of $x$), $x^- = (-x)^+ = \sup(-x, 0)$ (the negative part of $x$), $|x| = \sup(x, -x)$ (the absolute value of $x$); then $x = x^+ - x^-$ and $|x| = x^+ + x^-$; here, these two relations are equivalent to

The relation $x \leq y$ is equivalent to $\langle x^+ \leq y^+ \text{ and } x^- \geq y^- \rangle$. For any $x$ and $y$, the triangle inequality holds:

(1)

By the invariance of order under every homothety of ratio $> 0$,

(2)

In particular,

(3)

On the other hand, for $\lambda < 0$ we have $(\lambda x)^+ = (-\lambda x)^- = |\lambda| x^-$ and $(\lambda x)^- = (-\lambda x)^+ = |\lambda| x^+$; it follows that, for every $\lambda \in \mathbf{R}$ and every $x \in E$,

(4)

The invariance of order under translation shows that for all $z \in E$,

(5)

whence, in particular,

(6)

We have the relations

(7) $\inf(x, y) = -\sup(-x, -y) ,$

(8) $\sup(x, y) + \inf(x, y) = x + y .$

If $x, y, z$ are $\geq 0$ then (A, VI, §1, No. 12, Prop. 11)

(9) $\inf(x + y, z) \leq \inf(x, z) + \inf(y, z) .$

If A and B are two subsets of E each of which has a supremum, then A + B also has a supremum and

(10) $\sup(A + B) = \sup A + \sup B .$

Two elements $x, y$ of E are said to be alien$^3$ (to each other) if $\inf(|x|, |y|) = 0$; by (8), this relation is equivalent to $\sup(|x|, |y|) = |x| + |y|$, and also, by (6), to $||x| - |y|| = |x| + |y|$; 0 is the only element alien to itself; for every $x \in E$, $x^+$ and $x^-$ are alien and may be characterized as the only alien elements $y \geq 0, z \geq 0$ such that $x = y - z$. If $y$ is alien to $x$, then every $z \in E$ such that $|z| \leq |y|$ is also alien to $x$. If $y$ and $z$ are alien to $x$, then so is $|y| + |z|$, by the inequality (9); in particular, $n|y|$ is alien to $x$ for every integer $n > 0$, from which it follows that $\lambda y$ is alien to $x$ for every scalar $\lambda$, since there exists an integer $n$ such that $|\lambda| \leq n$, whence $|\lambda y| \leq n|y|$. If a subset A of E consists of elements alien to $x$ and if A has a supremum, then that supremum is also alien to $x$ (A, VI, §1, No. 12, Cor. of Prop. 13).

Finally, we have the decomposition lemma (A, VI, §1, No. 10, Th. 1):

If* $(x_i)_{i \in I}, (y_j)_{j \in J}$ are two finite sequences of elements $\geq 0$ of E such that $\sum_{i \in I} x_i = \sum_{j \in J} y_j$, then there exists a finite sequence $(z_{ij})_{(i,j) \in I \times J}$ of elements $\geq 0$ of E such that $x_i = \sum_{j \in J} z_{ij}$ for all $i \in I$, and $y_j = \sum_{i \in I} z_{ij}$ for all $j \in J$.

Cites 3

Cites VI § 1. Integration of vector-valued functions, Proposition 11, VI § 1, Theorem 1, VI § 1

Example

— 1) If A is any set, the space $\mathbf{R}^A$ of real-valued functions defined on A is fully lattice-ordered, the supremum in $\mathbf{R}^A$ of a family that is bounded above being its upper envelope (GT, IV, §5, No. 5).

2) Let F be any set; the space $\mathscr{B}(F)$ of bounded real-valued functions on F, equipped with the order structure induced by that of $\mathbf{R}^F$, is fully lattice-ordered. However, if F is a topological space, the space $\mathscr{C}(F)$ of continuous real-valued functions on F (equipped with the order structure induced by that of $\mathbf{R}^F$) is a Riesz space that is not in general fully lattice-ordered (cf. Exer. 13). Consider for example the case that $F = \mathbf{R}$; let I be the interval ]0, 1[, $\varphi_I$ the characteristic function of I, and let H be the set of continuous functions $x(t)$ such that $x \leq \varphi_I$; it is clear that H is bounded above in $\mathscr{C}(F)$. The function $\varphi_I$ is the upper envelope of the $x \in H$, but it is not their supremum in $\mathscr{C}(F)$, since $\varphi_I$ is lower semi-continuous but not continuous. Let us show that, in fact, H has no supremum in $\mathscr{C}(F)$; it suffices to prove that if $u$ is a continuous function such that $u \geq \varphi_I$, then there exists a continuous function $v \neq u$ such that $u \geq v \geq \varphi_I$. Now, $u(0) \geq 1$, therefore there exists a number $\alpha > 0$ such that $u(t) > 0$ for $-\alpha \leq t \leq 0$; if w is a continuous function that is zero outside of the interval ]$-\alpha, 0$ and is such that $0 < w(t) < u(t)$ on this interval, then the function $v = u - w$ meets the requirements.

Cites 1

Cites IV § 5. Measurable functions and sets

Example

In the space $\mathbf{R}^A$ of real-valued functions defined on a set $A$, the set of functions that are zero at all the points of a subset $M$ of $A$ is a band.